Hi,

I am hoping for some help on the following question:

A study has shown that the probability distribution of x, the number of customers in line (including the one being served, if any) at a checkout counter in a department store, is given by P (X=0) = 0.25, P(X=1) = 0.25, P(X=2) = 0.20, P(X=3)=0.20, and P(X_>4) =0.10. Considering a newly arriving customer to the checkout line.

a) what is the probability that this customer will not have to wait behind anyone
b) what is the probability that this customer will have to wait behind at least one customer
c) On average, behind how many other customers will the newly arriving customer have to wait?

For part a, I have the answer 0.25(P (X=0) = 0.25)
For part b my answer is 0.75 ((X=1) = 0.25 + P(X=2) = 0.20+ P(X=3)=0.20+ and P(X_>4) =0.10)

Is that correct because that seems to simple just to give the probabilities that they have already given us in the question.

For part c, I am not quite sure what they are asking, should I be finding the mean (average) by using the expected number of customers (0 to 4) and the probabilities for each and then summing them up?

Any help would be appreciated.

Thanks

a) Right.

b) Right, or simply, 1 - P(X = 0)

c) Yes, you are to find the mean, or expectation, which is the same.

It's (0 * 0.25) + (1 * 0.25) + (2 * 0.2) +... + (4 * 0.1)

However, I'm not sure for the last term for this is the first time there is no specific upper boundary...

The first two really ARE that easy. Good job! For part (b) however, you also could have approached the problem as follows: You know the total probability in adny situation always adds up to 1 (i.e. 100%). Therefore, P(X<1) + P(X>=1) = 1, or P(X>=1) = 1 - P(X<1) = 1 - 0.25 = 0.75. Your way was perfectly fine; you got the same answer. But in the future you may find you need to answer such questions without explicitly knowing all the probabilities for X=1, X=2, etc.

As for part (c), I think the question is poorly worded. The words "on average" do indeed indicate that you should find the mean (a.k.a. "average"). However, it's impossible to find the mean with the information supplied because of that darned P(X>=4) term. All it tells you is that 10% of the time there are 4 or more people in line. Maybe that means 5% of the time there are 4 people, and 5% of the time there are 5 people (but never 6 or more). After all, 4 and 5 are both >= 4. Or, on the other hand, maybe that means that 5% of the time there are 4 people, and 5% of the time there are 6 million people in line! Obviously that's unrealistic, but it still meets the criterion that X >= 4, so it's mathematically allowable. You can imagine how much that would skew the average! My point is simply to illustrate that you can't compute the mean with the information supplied.

What I think the question is actually asking (despite its poor word choice), is for the median number of people in line. That's the number for which 50% of the distribution falls below and the other 50% falls above. Do you know how to do that?

Ah, I was having a doubt about the last term and was offline for a while before I submitted my edit. I looked in my stats book for such an occurrence but didn't find anything, that might be the reason why :)

X should be treated as 4 here.

Can you explain why so? Because given the statement, we should also anticipate that x could be anything greater than 4 because of the 'greater' sign.